I need to solve some cubic equations, but we didn't study how to do so. So, please can you provide a tip in order to solve cubic equations. Without factoring it, because it's not all the time e=0. I want something like b^2-4ac for square equations. Ex: $x^2-x-2=0$

  1. Step1: we could do some factoring like $(x-1)(x+2)=0$
    Step2: then $x=1$ or $x=-2$ that's it.
  2. Step1: $b^2-4ac = 1-4(1)(-2) = 9$
    Step2: $x_1 = \frac{1-\sqrt9}2 = -4/2 = -2$ or $x_2= \frac{1+\sqrt9}2 = 2/2 = 1$

In short, I wanna something like the second one not the first, something automatically gives you the answer.

Thanks, In Advance.

  • $\begingroup$ google "roots of cubic equations" $\endgroup$ Commented Sep 23, 2015 at 11:06
  • $\begingroup$ Google it as he said and please write in latex, it is unreadable as iti s now. $\endgroup$ Commented Sep 23, 2015 at 11:06
  • $\begingroup$ @ZelosMalum, could you kindly refrain from biting the newcomers like that? It's perfectly clear which formulas he intends to write, and if you want them to look prettier, you can suggest an edit to TeXify them yourself. $\endgroup$ Commented Sep 23, 2015 at 11:10
  • $\begingroup$ @uniquesolution I did, I found in wiki how to find Delta. Yet, I didn't find how to find the exact roots $\endgroup$ Commented Sep 23, 2015 at 11:11
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    $\begingroup$ @AmineMarzouki: What most of us do is to code the formulas in MathJax (or LaTeX) format on a regular keyboard. See this tutorial on Meta for help on how to format various kinds of formulas -- but for a basic start, you can just enclose "ASCII math" in dollar signs: writing $x^2-x-2=0$ produces $x^2-x-2=0$. $\endgroup$ Commented Sep 23, 2015 at 11:20

1 Answer 1


There are explicit method for solving general cubic equations -- google for "Cardano's formula" -- but they are so complex (!) that they are close to useless for most practical purposes. Even when the roots are nice rational numbers, the general techniques tend to produce them as a maze of nested square and cube roots, which it is not easy to see equals the rational result.

What one does in practice is either to guess a rational root, factor it out, and solve the resulting quadratic, or to approximate the roots numerically using techniques like bisection or Newton-Raphson.

  • $\begingroup$ Thank you so much for your kind answer. I read about Cardano's formula method in wiki. I didn't understand it because I still didn't study complex numbers (I will this year) I'll search for bisection or Newton-Raphson. $\endgroup$ Commented Sep 23, 2015 at 11:20
  • $\begingroup$ @AmineMarzouki Don't be scared of complex numbers, they are very friendly and don't bite. They might do some roughhousing though :P $\endgroup$ Commented Sep 23, 2015 at 11:23
  • $\begingroup$ @AmineMarzouki: Actually, in the case where the equation has three real roots, Cardano's formula does not require complex numbers. It is still pretty horrible, though -- and since there's provably no such formula for fifth or higher degree equations, cubic seems to be a reasonable place in practice to cut one's losses and move to numerical methods, given that we have to do it sooner or later anyway. $\endgroup$ Commented Sep 23, 2015 at 11:24
  • $\begingroup$ @ZelosMalum Haha, I hope so $\endgroup$ Commented Sep 23, 2015 at 11:25
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    $\begingroup$ Henning, you got a word backwards, it is in the case of three irrational real roots that there must be complex numbers and no way to improve the situation using algebraic methods, en.wikipedia.org/wiki/Casus_irreducibilis $\endgroup$
    – Will Jagy
    Commented Sep 23, 2015 at 19:58

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